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3 years ago

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A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the li

ght leave the water?

1 answer:

suter [353]3 years ago

3 0

Answer:

45.93°

Explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is $n_1=1.33$

Refractive index of the air that is $n_2=1$ (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law $n_1sini=n_2sinr$

$1.33sin32.7=1\times sinr$

$sinr=0.7185$

r =45.93°

So the light leave the water at an angle of 45.93°

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1 year ago

What is electrons ? explain it function<br>

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3 years ago

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What is the average speed of an object that travels 6 meters in 2 seconds and then travels 3 meters in 1 second?

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3 m/s

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2 years ago

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

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Answer:

Maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

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For this we have to find the angular frequency

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3 years ago

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

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so

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

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